
First, from the proof of Theorem 2 in [1]}, we have \(\sum _{i,u}\lambda _{i,u}{\rm Q}_{i,u}\)  in (REF ) rewritten as
\(&\sum _{i,u}\lambda _{i,u}{\rm Q}_{i,u} = \alpha \beta \sum _{i,u}\!{\bf w}_{i,u}^H{\rm diag}\left({\bf H}_i{\Lambda }{\bf H}_i^H\!\right)\!{\bf w}_{i,u}.\) 
